$\dfrac{ -i + 6j }{ 3 } = \dfrac{ -6i + 9k }{ -8 }$ Solve for $i$.
Solution: Multiply both sides by the left denominator. $\dfrac{ -i + 6j }{ {3} } = \dfrac{ -6i + 9k }{ -8 }$ ${3} \cdot \dfrac{ -i + 6j }{ {3} } = {3} \cdot \dfrac{ -6i + 9k }{ -8 }$ $-i + 6j = {3} \cdot \dfrac { -6i + 9k }{ -8 }$ Multiply both sides by the right denominator. $-i + 6j = 3 \cdot \dfrac{ -6i + 9k }{ -{8} }$ $-{8} \cdot \left( -i + 6j \right) = -{8} \cdot 3 \cdot \dfrac{ -6i + 9k }{ -{8} }$ $-{8} \cdot \left( -i + 6j \right) = 3 \cdot \left( -6i + 9k \right)$ Distribute both sides $-{8} \cdot \left( -i + 6j \right) = {3} \cdot \left( -6i + 9k \right)$ ${8}i - {48}j = -{18}i + {27}k$ Combine $i$ terms on the left. ${8i} - 48j = -{18i} + 27k$ ${26i} - 48j = 27k$ Move the $j$ term to the right. $26i - {48j} = 27k$ $26i = 27k + {48j}$ Isolate $i$ by dividing both sides by its coefficient. ${26}i = 27k + 48j$ $i = \dfrac{ 27k + 48j }{ {26} }$